-0.016 738 891 601 562 496 524 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 524(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.016 738 891 601 562 496 524(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.016 738 891 601 562 496 524| = 0.016 738 891 601 562 496 524


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.016 738 891 601 562 496 524.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.016 738 891 601 562 496 524 × 2 = 0 + 0.033 477 783 203 124 993 048;
  • 2) 0.033 477 783 203 124 993 048 × 2 = 0 + 0.066 955 566 406 249 986 096;
  • 3) 0.066 955 566 406 249 986 096 × 2 = 0 + 0.133 911 132 812 499 972 192;
  • 4) 0.133 911 132 812 499 972 192 × 2 = 0 + 0.267 822 265 624 999 944 384;
  • 5) 0.267 822 265 624 999 944 384 × 2 = 0 + 0.535 644 531 249 999 888 768;
  • 6) 0.535 644 531 249 999 888 768 × 2 = 1 + 0.071 289 062 499 999 777 536;
  • 7) 0.071 289 062 499 999 777 536 × 2 = 0 + 0.142 578 124 999 999 555 072;
  • 8) 0.142 578 124 999 999 555 072 × 2 = 0 + 0.285 156 249 999 999 110 144;
  • 9) 0.285 156 249 999 999 110 144 × 2 = 0 + 0.570 312 499 999 998 220 288;
  • 10) 0.570 312 499 999 998 220 288 × 2 = 1 + 0.140 624 999 999 996 440 576;
  • 11) 0.140 624 999 999 996 440 576 × 2 = 0 + 0.281 249 999 999 992 881 152;
  • 12) 0.281 249 999 999 992 881 152 × 2 = 0 + 0.562 499 999 999 985 762 304;
  • 13) 0.562 499 999 999 985 762 304 × 2 = 1 + 0.124 999 999 999 971 524 608;
  • 14) 0.124 999 999 999 971 524 608 × 2 = 0 + 0.249 999 999 999 943 049 216;
  • 15) 0.249 999 999 999 943 049 216 × 2 = 0 + 0.499 999 999 999 886 098 432;
  • 16) 0.499 999 999 999 886 098 432 × 2 = 0 + 0.999 999 999 999 772 196 864;
  • 17) 0.999 999 999 999 772 196 864 × 2 = 1 + 0.999 999 999 999 544 393 728;
  • 18) 0.999 999 999 999 544 393 728 × 2 = 1 + 0.999 999 999 999 088 787 456;
  • 19) 0.999 999 999 999 088 787 456 × 2 = 1 + 0.999 999 999 998 177 574 912;
  • 20) 0.999 999 999 998 177 574 912 × 2 = 1 + 0.999 999 999 996 355 149 824;
  • 21) 0.999 999 999 996 355 149 824 × 2 = 1 + 0.999 999 999 992 710 299 648;
  • 22) 0.999 999 999 992 710 299 648 × 2 = 1 + 0.999 999 999 985 420 599 296;
  • 23) 0.999 999 999 985 420 599 296 × 2 = 1 + 0.999 999 999 970 841 198 592;
  • 24) 0.999 999 999 970 841 198 592 × 2 = 1 + 0.999 999 999 941 682 397 184;
  • 25) 0.999 999 999 941 682 397 184 × 2 = 1 + 0.999 999 999 883 364 794 368;
  • 26) 0.999 999 999 883 364 794 368 × 2 = 1 + 0.999 999 999 766 729 588 736;
  • 27) 0.999 999 999 766 729 588 736 × 2 = 1 + 0.999 999 999 533 459 177 472;
  • 28) 0.999 999 999 533 459 177 472 × 2 = 1 + 0.999 999 999 066 918 354 944;
  • 29) 0.999 999 999 066 918 354 944 × 2 = 1 + 0.999 999 998 133 836 709 888;
  • 30) 0.999 999 998 133 836 709 888 × 2 = 1 + 0.999 999 996 267 673 419 776;
  • 31) 0.999 999 996 267 673 419 776 × 2 = 1 + 0.999 999 992 535 346 839 552;
  • 32) 0.999 999 992 535 346 839 552 × 2 = 1 + 0.999 999 985 070 693 679 104;
  • 33) 0.999 999 985 070 693 679 104 × 2 = 1 + 0.999 999 970 141 387 358 208;
  • 34) 0.999 999 970 141 387 358 208 × 2 = 1 + 0.999 999 940 282 774 716 416;
  • 35) 0.999 999 940 282 774 716 416 × 2 = 1 + 0.999 999 880 565 549 432 832;
  • 36) 0.999 999 880 565 549 432 832 × 2 = 1 + 0.999 999 761 131 098 865 664;
  • 37) 0.999 999 761 131 098 865 664 × 2 = 1 + 0.999 999 522 262 197 731 328;
  • 38) 0.999 999 522 262 197 731 328 × 2 = 1 + 0.999 999 044 524 395 462 656;
  • 39) 0.999 999 044 524 395 462 656 × 2 = 1 + 0.999 998 089 048 790 925 312;
  • 40) 0.999 998 089 048 790 925 312 × 2 = 1 + 0.999 996 178 097 581 850 624;
  • 41) 0.999 996 178 097 581 850 624 × 2 = 1 + 0.999 992 356 195 163 701 248;
  • 42) 0.999 992 356 195 163 701 248 × 2 = 1 + 0.999 984 712 390 327 402 496;
  • 43) 0.999 984 712 390 327 402 496 × 2 = 1 + 0.999 969 424 780 654 804 992;
  • 44) 0.999 969 424 780 654 804 992 × 2 = 1 + 0.999 938 849 561 309 609 984;
  • 45) 0.999 938 849 561 309 609 984 × 2 = 1 + 0.999 877 699 122 619 219 968;
  • 46) 0.999 877 699 122 619 219 968 × 2 = 1 + 0.999 755 398 245 238 439 936;
  • 47) 0.999 755 398 245 238 439 936 × 2 = 1 + 0.999 510 796 490 476 879 872;
  • 48) 0.999 510 796 490 476 879 872 × 2 = 1 + 0.999 021 592 980 953 759 744;
  • 49) 0.999 021 592 980 953 759 744 × 2 = 1 + 0.998 043 185 961 907 519 488;
  • 50) 0.998 043 185 961 907 519 488 × 2 = 1 + 0.996 086 371 923 815 038 976;
  • 51) 0.996 086 371 923 815 038 976 × 2 = 1 + 0.992 172 743 847 630 077 952;
  • 52) 0.992 172 743 847 630 077 952 × 2 = 1 + 0.984 345 487 695 260 155 904;
  • 53) 0.984 345 487 695 260 155 904 × 2 = 1 + 0.968 690 975 390 520 311 808;
  • 54) 0.968 690 975 390 520 311 808 × 2 = 1 + 0.937 381 950 781 040 623 616;
  • 55) 0.937 381 950 781 040 623 616 × 2 = 1 + 0.874 763 901 562 081 247 232;
  • 56) 0.874 763 901 562 081 247 232 × 2 = 1 + 0.749 527 803 124 162 494 464;
  • 57) 0.749 527 803 124 162 494 464 × 2 = 1 + 0.499 055 606 248 324 988 928;
  • 58) 0.499 055 606 248 324 988 928 × 2 = 0 + 0.998 111 212 496 649 977 856;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.016 738 891 601 562 496 524(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 524(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 524(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(2) × 2-6


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1017(10) =

011 1111 1001(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


Decimal number -0.016 738 891 601 562 496 524 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100